Biomolecules Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Biomolecules — carbohydrates, amino acids, vitamins, and proteins — with step-by-step solutions.
A curated set of JEE Main 2026 previous year questions from Biomolecules, each with a concise worked solution.
Solutions are AI-generated and pending review.
Solution
Recall the standard chemical names of the B-complex vitamins and vitamin C:
- Vitamin $\text{B}_1$ $=$ Thiamine $\Rightarrow$ III
- Vitamin $\text{B}_2$ $=$ Riboflavin $\Rightarrow$ IV
- Vitamin $\text{B}_6$ $=$ Pyridoxine $\Rightarrow$ I
- Vitamin C $=$ Ascorbic acid $\Rightarrow$ II
So the matching is A-III, B-IV, C-I, D-II.
Answer: C
Solution
The one-letter codes: T = Threonine and Y = Tyrosine.
The thyroid hormone thyroxine ($\text{T}_4$) is synthesised by iodination of the aromatic ring of tyrosine (Y) residues in thyroglobulin. Insulin is a peptide hormone, not an iodinated amino acid derivative.
Hence the correct pair is amino acid $=$ Y (tyrosine), hormone $=$ thyroxine.
Answer: C
Solution
Match each amino acid to its one-letter code and nutritional class:
- Arginine $\to$ code R, essential (semi-essential) $\Rightarrow$ II (R / Essential)
- Aspartic acid $\to$ code D, non-essential $\Rightarrow$ I (D / Non-essential)
- Lysine $\to$ code K, essential $\Rightarrow$ IV (K / Essential)
- Glutamic acid $\to$ code E, non-essential $\Rightarrow$ III (E / Non-essential)
So A-II, B-I, C-IV, D-III.
Answer: A
Solution
An aldotetrose is $\text{CHO}-\text{CHOH}-\text{CHOH}-\text{CH}_2\text{OH}$. Oxidation with conc. $\text{HNO}_3$ converts both the terminal $\text{CHO}$ and $\text{CH}_2\text{OH}$ into $-\text{COOH}$, giving a tartaric acid (2,3-dihydroxybutanedioic acid).
D-series requirement: the reference carbon C3 (lowest chiral centre) must have its $\text{OH}$ on the right. This rules out any structure with C3-$\text{OH}$ on the left.
The two D-aldotetroses are:
- D-erythrose — C2-$\text{OH}$ right, C3-$\text{OH}$ right (both same side).
- D-threose — C2-$\text{OH}$ left, C3-$\text{OH}$ right (opposite sides).
The two $\text{OH}$ groups end up on the same side, giving a molecule with an internal plane of symmetry $\Rightarrow$ optically inactive. D-threose instead gives an optically active tartaric acid.
Therefore the required structure has C2-$\text{OH}$ right and C3-$\text{OH}$ right (D-erythrose), which is option C.
Answer: C
Solution
Evaluate each statement:
- A — True. Glucose exists as $\alpha$- and $\beta$-anomers.
- B — True. The anomers differ only in configuration at the anomeric carbon C-1 of the cyclic hemiacetal.
- C — False. $\alpha$-D-glucose melts at $\approx 146\,^\circ\text{C}$ and $\beta$-D-glucose at $\approx 150\,^\circ\text{C}$, so $\alpha < \beta$.
- D — False. The specific rotations are swapped: $\alpha$-anomer $= +111^\circ$ (≈ $+112^\circ$) and $\beta$-anomer $= +19^\circ$.
- E — True. $\alpha$-glucose crystallises from saturated solution at $303\ \text{K}$ and $\beta$-glucose above $371\ \text{K}$.
Correct statements: A, B and E.
Answer: D
Solution
Assess each statement about tertiary structure:
- Fibrous or globular — correct; tertiary structure gives rise to these two broad shapes.
- Stabilising forces (H-bonding, disulphide links, van der Waals, electrostatic) — correct.
- “Structure remains intact when exposed to pH changes” — incorrect. A change in pH disrupts ionic/hydrogen bonding and causes denaturation, unfolding the tertiary structure.
- Linear chain $\to$ secondary $\to$ tertiary — correct description of the folding hierarchy.
The false statement is the one claiming pH-stability.
Answer: C
Solution
Link each deficiency disease to the vitamin whose lack causes it:
- Scurvy $\leftarrow$ deficiency of Ascorbic acid (Vit C) $\Rightarrow$ III
- Convulsions $\leftarrow$ deficiency of Pyridoxine ($\text{B}_6$) $\Rightarrow$ I
- Cheilosis $\leftarrow$ deficiency of Riboflavin ($\text{B}_2$) $\Rightarrow$ IV
- Xerophthalmia $\leftarrow$ deficiency of Vitamin A $\Rightarrow$ II
So A-III, B-I, C-IV, D-II.
Answer: C
Solution
Identify the side-chain functional group and its characteristic test:
- Glutamine — side chain has an amide ($-\text{CONH}_2$) $\Rightarrow$ Hofmann bromamide degradation (IV)
- Lysine — side chain has a primary amine ($-\text{NH}_2$) $\Rightarrow$ Hinsberg’s test (I)
- Tyrosine — side chain has a phenolic $-\text{OH}$ $\Rightarrow$ neutral $\text{FeCl}_3$ test (II)
- Serine — side chain has a primary alcohol ($-\text{CH}_2\text{OH}$) $\Rightarrow$ ceric ammonium nitrate test (III)
So A-IV, B-I, C-II, D-III.
Answer: B
Solution
Check each statement:
- “All monosaccharides are reducing sugars.” — correct (all have a free anomeric $-\text{OH}$/carbonyl).
- “The monosaccharide units obtained from hydrolysis of oligosaccharides are always the same.” — incorrect. Sucrose $\to$ glucose $+$ fructose; lactose $\to$ glucose $+$ galactose — the units can be different.
- “Starch and cellulose are polysaccharides of more than ten monosaccharide units.” — correct.
- “Open chain and cyclic structures co-exist at equilibrium, as in $D-(+)$-glucose.” — correct (mutarotation).
The incorrect statement is the second one.
Answer: B
Solution
A violet/coloured complex with neutral $\text{FeCl}_3$ is characteristic of a phenolic $-\text{OH}$ group.
- Threonine, Serine — bear aliphatic ($-\text{CHOH}$ / $-\text{CH}_2\text{OH}$) hydroxyls, no colour.
- Cysteine — bears a thiol ($-\text{SH}$), no violet colour.
- Tyrosine — bears a phenol ring $\Rightarrow$ gives the violet Fe–phenolate complex.
Answer: C